Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.4 First-Order Linear Equations - Exercises - Page 524: 4


$$y=e^{ 2x} $$

Work Step by Step

This is a linear equation and has the integrating factor as follows $$\alpha(x)= e^{\int P(x)dx}=e^{\int (-1)dx}=e^{-x}$$ Now the general solution is \begin{align} y&=\alpha^{-1}(x)\left( \int \alpha(x)Q(x)dx +C\right)\\ &=e^x\left( \int e^{x}dx +C\right)\\ &=e^x\left( e^{x} +C\right) \end{align} At $x=0$, $y=1$, and we have $C=0$ Thus, the general solution is $$y=e^{ 2x} $$
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